the answer for your question is yes. Note that, all you need to prove is the following equality
$$
\int_{E} X\ dP
=
\int_{E} \left[\int_{\Omega} X\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}} \qquad \qquad (1).
$$
for all $E\in\mathcal{N}$.
The simplest way to do this is start from characteristic functions and then use linearity and finally the powerfull approximation theorems of the measure theory.
Let us start with the case $X=1_{A}$ where $A\in \mathcal{F}$. In this case the above equality holds as long as
$$
P(A\cap E)=\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}}.
$$
But this is certainly true and we can prove this manipulating the rhs. Indeed
$$
\int_{\Omega} \left[\int_{\Omega} 1_{A\cap E}\ dP(\cdot|\mathcal{N})(\omega) \right]
dP|_{\mathcal{N}}
=
\int_{\Omega} \mathbb{E}(A\cap E|\mathcal{N})
dP|_{\mathcal{N}}
=
\int_{E} \mathbb{E}(A|\mathcal{N})
dP|_{\mathcal{N}},
$$
where we used in the last equality that $E$ is $\mathcal{N}$ measurable.
But the right hand side above is by the properties of conditional expectation equals to $P(A\cap E)$. So the identity (1) holds for any characteristic function and for all $E\in\mathcal{N}$.
By linearity of the integral (1) is certainly true if $X$ is any simple function. Let $X$ be any positive $\mathcal{F}-$measurable function. We know that there is a sequence of simple functions $\varphi_n \uparrow X$, here the up arrow means monotone convergence. Since at this point is clear for us that
$$
\begin{eqnarray}
\int_{E} \varphi_n\ dP
&=&
\int_{E} \left[\int_{\Omega} \varphi_n\ dP(\cdot|\mathcal{N})(\omega) \right] dP|_{\mathcal{N}}
\\
&=& \int_{E} \mathbb{E}(\varphi_n|\mathcal{N}) dP|_{\mathcal{N}},
\end{eqnarray}
$$
we can finish the proof by invoking the Monotone Convergence Theorem.
Edition. I corrected the error pointed out by Didier Piau.
Addition. Theorem. Let $(\Omega,\mathcal{F},\mu)$ be a $\sigma$-finite measure space, $\mathcal{A}$ a sub-$\sigma$-algebra of $\mathcal{B}$, and $\nu=\mu|_{\mathcal{A}}$. If $f\in L^1(\mu)$, there exist $g\in L^1(\nu)$ (thus $g$ is $\mathcal{A}$-measurable) such that
$$
\int_{E} f\ d\mu =\int_{E} g\ d\nu
$$
for all $E\in A$; if $g'$ is another such function then $g=g'$ $\nu$-a.e. (Note that $g$ is the conditional expectation of $f$ on $\mathcal{A}$. (Folland)
Let $\cal H$ be the $\sigma$-algebra generated by ${\Bbb E} (X\mid\mathcal G)$.
As you say, since ${\Bbb E} (X\mid\mathcal G)$ is $\mathcal G$-measurable, $\cal H$ is contained in $\mathcal G$. It doesn't have to be contained in $\sigma(X)$. For a simple counterexample, let the probability space be $\Omega=\{1,2,3,4\}$ with all subsets measurable and the uniform probability measure, and let $${\mathcal G}=\{\emptyset,\{1,2\},\{3\},\{4\},\{3,4\},\{1,2,3\},\{1,2,4\},\Omega\},$$ $$X(1)=1, X(2)=X(3)=X(4)=0.$$ Then
$$
{\Bbb E} (X\mid\mathcal G)(1)={\Bbb E} (X\mid\mathcal G)(2)=\frac12,
\ \ \ \
{\Bbb E} (X\mid\mathcal G)(3)={\Bbb E} (X\mid\mathcal G)(4)=0$$
so ${\cal H}=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$ is not contained in $\sigma(X)=\{\emptyset,\{1\},\{2,3,4\},\Omega\}$.
Since $\cal H$ does not have to be contained in $\sigma(X)$, it need not equal ${\cal G}\cap\sigma(X)$. The example above shows that it also does not have to equal $\cal G$.
Addendum: As Byron Schmuland says below, if $X$ is an ${\cal F}/{\cal B}({\Bbb R})$-measurable real-valued function, then $\cal H$ is countably generated as it's the pullback of the countably generated $\sigma$-algebra ${\cal B}(\Bbb R)$ under ${\Bbb E} (X\mid\mathcal G)^ {-1}$. Also, any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\cal H$ for some real-valued $X$. (${\Bbb E} (X\mid\mathcal G)$ is only defined up to equality on a set of measure $1$, so to be precise you should say that any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\sigma({\Bbb E}(X\mid\mathcal G))$ for some real-valued $X$ and some version of ${\Bbb E}(X\mid \mathcal G)$.)
Best Answer
The property doesn't hold when $\Omega=\{a,b,c\}$, $\cal F:= 2^\Omega$, $\cal G:=\{\emptyset,\{a,b\},\{c\},\Omega\}$, $\cal H:=\{\emptyset,\{a\},\{b,c\},\Omega\}$ and $X:=\chi_{\{b\}}$.